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Find the number of ordered quadruples $(a,b,c,d)$ of nonnegative real numbers such that \begin{align*} a^2 + b^2 + c^2 + d^2 &= 4, \\ (a + b + c + d)(a^3 + b^3 + c^3 + d^3) &= 16. \end{align*}

I have 21 as my answer since (1,1,1,1), (0,0,0,2), and $(0,0\sqrt{2},\sqrt{2})$ and found all the combinations possible with (0,0,0,2) and $(0,0\sqrt{2},\sqrt{2})$,then adding up all the possible combinations and getting 21. However this answer is incorrect. What did I do wrong?

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    $\begingroup$ $(a, b,c,d)=(0,0,2,2)$ does not satisfy the second equation. $\endgroup$ – Dietrich Burde Apr 25 at 18:13
  • $\begingroup$ How is this a Cauchy-Schwarz inequality question? $\endgroup$ – kccu Apr 25 at 18:20
  • $\begingroup$ He meant, that on can solve this system using the Cauchy-Schwarz inequality. $\endgroup$ – Dr. Sonnhard Graubner Apr 25 at 18:27
  • $\begingroup$ @Dr.SonnhardGraubner How so? $\endgroup$ – kccu Apr 25 at 18:48
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    $\begingroup$ @Dr. Sonnhard Graubner please do not assume my gender. That is rude. $\endgroup$ – sumi Apr 25 at 18:59
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$$a^2 + b^2 + c^2 + d^2 = 4$$

$$(a + b + c + d)(a^3 + b^3 + c^3 + d^3) = 16$$

By the C.S. inequality $$(a + b + c + d)(a^3 + b^3 + c^3 + d^3) \ge (a^2 + b^2 + c^2 + d^2)^2$$

In this case they are equal, and the equality holds only when: $$\frac{a}{a^2} = \frac{b}{b^2} = \frac{c}{c^2} = \frac{d}{d^2}$$ $$\implies a = b = c = d = 1$$

Note that if any of $\{a,b,c,d\}$ are $0$, the ratios will not exist. We need to consider them on a case-by-case basis:

  1. One of them, say $a=0$: In this case, $$b=c=d=\frac{2}{\sqrt{3}}$$
  2. Two of them are zero, say $a=b=0$. Then, $$b = c = \sqrt{2}$$
  3. Three of them say $a=b=c=0$. Then $d=2$

Hence the solutions are: $$\boxed{(a,b,c,d) = (1,1,1,1),\ (0, \frac{2}{\sqrt{3}},\ \frac{2}{\sqrt{3}}, \frac{2}{\sqrt{3}}),\ (0, 0, \sqrt{2}, \sqrt{2}),\ (0,0,0,2)}$$

Number of possibilities:

  1. $(1,1,1,1) \implies ^4C_4 = 1$
  2. $(0, \frac{2}{\sqrt{3}}, \frac{2}{\sqrt{3}}, \frac{2}{\sqrt{3}}) \implies ^4C_1 = 4$
  3. $(0, 0, \sqrt{2}, \sqrt{2}) \implies ^4C_2 = 6$
  4. $(0,0,0,2) \implies ^4C_1 = 4$

Hence the total number of cases seems to be $\boxed{15}$

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  • $\begingroup$ That's not true, as you can see $(0,0,\sqrt{2},\sqrt{2})$ is a solution. $\endgroup$ – kccu Apr 25 at 18:52
  • $\begingroup$ @kccu (-1, -1, -1, -1) is a solution as per this argument but the problem states that the numbers are non-negative. I need to check how CS works for zeros $\endgroup$ – user1952500 Apr 25 at 18:53
  • $\begingroup$ If some of the terms are zero, then you need to apply C-S with fewer terms (e.g., $(a+b+c)(a^3+b^3+c^3)\geq (a^2+b^2+c^2)^2$ if $d=0$). You will find the additional solutions $(0,0,0,2)$, $(0,0,\sqrt{2},\sqrt{2})$, and $(0, \frac{2}{\sqrt{3}},\frac{2}{\sqrt{3}},\frac{2}{\sqrt{3}})$, as well as all permutations of those. $\endgroup$ – kccu Apr 25 at 18:57
  • $\begingroup$ @kccu I was just now writing down the exact same thing. Please take a look $\endgroup$ – user1952500 Apr 25 at 19:00
  • $\begingroup$ Don't forget permutations of those (e.g., $(\sqrt{2},0,\sqrt{2},0)$, $(0,2,0,0)$, etc.). $\endgroup$ – kccu Apr 25 at 19:17

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